Given the hyperbola x ＾ 2-y ＾ 2 / 2 = 1. The straight line passing through point a (2,1) intersects with the hyperbola at P1 and P2, find the trajectory equation of the midpoint P of line p1p2

Given the hyperbola x ＾ 2-y ＾ 2 / 2 = 1. The straight line passing through point a (2,1) intersects with the hyperbola at P1 and P2, find the trajectory equation of the midpoint P of line p1p2

Let P1 (x1, Y1), P2 (X2, Y2) and the midpoint P (x, y) of line p1p2, then x1 ＾ 2-y1 ＾ 2 / 2 = 1,2 ＾ 2-y2 ＾ 2 / 2 = 1, subtracting the two expressions to get: (x1 + x2) (x1-x2) - (Y1 + Y2) (y1-y2) / 2 = 0, ∵ X1 + x2 = 2x, Y1 + y2 = 2Y, ＾ 2x (x1-x2) - 2Y (y1-y2) / 2 = 0, (y1-y2) / (x1-x2) = 2x / y

Through the point (2, - 1), make a straight line intersection hyperbola 2x ^ 2-y ^ 2 = 2 at P and Q, and find the trajectory equation of the midpoint m of the line PQ

The linear slope is k, P (x1, Y1), q (X2, Y2), m (m, n) 2x1 ^ 2-y1 ^ 2 = 21 formula 2x2 ^ 2-y2 ^ 2 = 22 Formula 1 minus 2 formula, 2 (x1 ^ 2-x2 ^ 2) = Y1 ^ 2-y2 ^ 22 [(x1 + x2) / (Y1 + Y2)] = (y1-y2) / (x1-x2) 2m / N = K and K = (n + 1) / (m-2) then the result can be obtained

In triangle ABC, ad is the middle line on the side of BC, e is the middle point of AD, and passing through point a is the intersection of parallel lines of BC
&In triangle ABC, ad is the middle line on the side of BC, e is the middle point of AD, passing through point a as the parallel line of BC, intersecting the extension line of be at point F, connecting cf. & nbsp; if AB is perpendicular to AC, try to judge the shape of quadrilateral adcf, and prove your conclusion

1. ∵ ad is the middle line on the edge of BC, e is the middle point of AD, namely CD = af2, ∵ AF = CD, AE = de ∵ AF ∥ BC ∥ f = EBD, ∵ FAE = BDE ≌ AFE ≌ DBE (AAS) ∵ AF = BD = CD, that is, CD = af2,